I'm hoping to get some guidance and/or confirmation on the following problem. It is basically a combination of two results I already have, but I'm not totally sure I've got this one.
Consider the boundary value problem in $[0,T] \times \mathbb{R}$:
$\frac{\partial F}{\partial t}+2x\frac{\partial F}{\partial x}+8x^2\frac{\partial^2 F}{\partial x^2}+(\ln x)^2=0$ , $t>0$ , $x \in \mathbb{R}$
$F(T,x)=\ln (x^6)$
Find a stochastic representation for the solution in terms of the expected value of some function of the diffusion process which is a solution of the SDE
$dX_s=2X_sds+4X_sdBs$
$X_t=x$
My attempt:
So, we know by Ito that $dF(s,X_s)=\frac{\partial F}{\partial s}ds+\frac{\partial F}{\partial x}dX_s+\frac{1}{2}\frac{\partial^2 F}{\partial x^2}(dX_s)^2$ =
$\frac{\partial F}{\partial s}ds+\frac{\partial F}{\partial x}(2X_sds+4X_sdBs)+\frac{1}{2}\frac{\partial^2 F}{\partial x^2}8X_s^2ds$ = $(\frac{\partial F}{\partial s}+\frac{\partial F}{\partial x}2X_s+\frac{1}{2}\frac{\partial^2 F}{\partial x^2}8X_s^2)ds+\frac{\partial F}{\partial x}4X_sdBs$
But note that the entire term in parenthesis now equals $-(\ln x)^2$ by the original problem =>
$dF(s,X_s)=-(\ln x)^2ds+4X_s\frac{\partial F}{\partial x}dBs$ , integrating from t to T yields =>
$\int_t^TdF(s,X_s)=-\int_t^T(\ln x)^2ds+\int_t^T4X_s\frac{\partial f}{\partial x}dBs$
Now, note that the l.h.s. of the above = $F(T,X_T)-F(t,X_t)$, so =>
$E_{t,x}[F(T,X_T)]=E_{t,x}[F(t,X_t)]-\int_t^TE_{t,x}[(\ln x)^2]ds+0$
The zero is due to expectation of a stochastic integral, and also noting the terminal condition specified and also the fact that $F(t,X_t)$ is not really random, we have
$E_{t,x}[ln(x^6)]=F(t,x)-\int_t^TE_{t,x}[(\ln x)^2]ds$ <=> $F(t,x)=E_{t,x}[ln(x^6)]+\int_t^TE_{t,x}[(\ln x)^2]ds$